# Homework Help: Calculating Current using Kirkoff's Rules

1. Oct 13, 2012

### MasterVivi

1. The problem statement, all variables and given/known data
Apply the loop rule to two of the three loops to calculate I1,I2,and I3.

Picture of DC circuit involved

2. Relevant equations

V=IR

3. The attempt at a solution
L1=2v+2I2-4I1=0 (simplified)
and
L3=4-4I1-5I3=0

But if you work through the algebra you'd get
I2=1
which would then make
I1=1

which does not work for the junction rule of
I3=I1-I2 cause that is 0.

The rules still confuse me when look at some of the resistors on the right side I can tell which way current flows, obviously did it wrong though, any help would be appreciated.

2. Oct 13, 2012

### Spinnor

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3. Oct 13, 2012

### OmCheeto

1Ω + 3Ω ≠ 2Ω

4. Oct 13, 2012

### MasterVivi

Okay, re-attempted using loop one and loop two, does, this seem more accurate?

L1=12V-I1(2)+I2(1)-10V+I2(3)-I1(1)=0
=>
L1=2V-4I1+4I2=0

L2=-3(I2)+10V-I2(1)-I3(1)-8V-I3(2)=0
=>
L2=2V-4I2-5I3=0
=>
L2=2-4I2-5(I1-I2)=0
=>
L2=2-4I2-5I1+5I2=0
=>
L2=2-5I1+I2=0

So from there
I2=5I1-2

Finally plugging it all through
I1=I2+(1/2)[From the first loop equation]
=>
I1=5I1-2+(1/2)
=>
I1=(3/8)

if thats true

I2=5(3/8)-2=(-1/8) [Is a negative current okay???]

and from there

I3=I1-I2=(3/8)-(-1/8)=1/2

The negative current for I2 feels wrong, did I mess up any of the loop equation or any algebra along the way?

5. Oct 13, 2012

### frogjg2003

For the second loop, it should be +5I2, not -5I2
As for there being a negative current, it just means it's flowing in the opposite direction from what you set it up as in the problem.

6. Oct 14, 2012

### MasterVivi

Okay, think I got it

You're right it should be +5I3 because it's with the current, just like I2 works out in the first loop equation

L1=2-4I1+4I2=0

Using this to solve for I1

I1=I2+(1/2)

Now I did this wrong, so going back through the derivation of this loop equation

L2=-3I2+10-I2+2I3+I3-8+2I3=0
=>
2-4I2+5I3=0
=>
2-4I2+5(I1-I2)=0
=>
L2=1+5I1-9I2=0

Using that to solve for I2

I2=(5/9)I1+(2/9)

Solving for the numerical value of I1
I1=((5/9)I1+(2/9))+(1/2)
=>
I1=(5/9)I1+(13/18)
=>
(4/9)I1=(13/18)
=>
I1=(13/8)

Solving for the numerical Value of I2

I2=(5/9)(13/8)+(2/9)
=>
I2=(9/8)

Which Finally implies

I3=I1-I2
=>
I3=(13/8)-(9/8)=(1/2)

Aye?Nay?

7. Oct 15, 2012

### frogjg2003

correct
quick note: you had I1=I2+1/2, so you could have easily figured out I3 right then and there. I3 = I1-I2 = I2+1/2-I2 = 1/2

8. Oct 15, 2012

### OmCheeto

really?

L2=1+5I1-9I2=0
i1=13/8
i2=9/8
ergo:
1+5*(13/8)-9*(9/8)=0
8/8+65/8-81/8=0
73/8-81/8=0
-8/8=0
-1=0 ?

Don't all 4 equations have to come out correctly?